Acharged particle moves at 2.5 × 104 m/s at an angle of 25° to a magnetic field that has a field strength of 8.1 × 10–2 t. if the magnetic force is 7.5 × 10–2 n, what is the magnitude of the charge?

the magnitude of charge=q=8.76 x 10⁻⁵C

Explanation:

the magnetic force Fm is given by

Fm= q V B sinθ

q= charge

v= velocity= 2.5 x 10⁴ m/s

B= magnetic field strength= 8.1 x 10⁻²T

Fm= magnetic force= 7.5 x 10⁻² N

θ=25°

so 7.5 x 10⁻² =q (2.5 x 10⁴ ) (8.1 x 10⁻²) sin25

q=8.76 x 10⁻⁵C

answer is C on edg

Explanation:

1.195000

explanation:

total mass=0.0117+0.0137+0.0881+0.0111

total mass= 0.01172kg

total velocity = 0.283-0.743+0.871-0.799+0.155+0.637+0.301=0.102

momentum=mv=0.1172*0.102=1.195000//

80cm/s.

24cm/s.

53.3cm/s.